# Top Chern Class = Euler Characteristic

Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality

$\displaystyle \int_X c_n(\mathcal{T}_X) = \chi(X^{\mathrm{an}})$,

i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance.

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I think the difficulty you're having finding a reference is you're asking the question in too local a setting -- Chern classes are for complex bundles over spaces. In that setting, Wikipedia gives you the reference: en.wikipedia.org/wiki/Chern_class –  Ryan Budney Aug 23 '11 at 0:22
You mean the original paper by Chern? I thought that since 1945, it might have been elaborated in a more algebraic setting and greater detail, which is what I was rather looking for. –  Jesko Hüttenhain Aug 23 '11 at 0:51
The top Chern class is the obstruction to a section of the bundle. In the case of the tangent bundle, the Poincare-Hopf Index Theorem (see for example Milnor and Stasheff, or the textbook Guillemin and Pollack) tells you evaluation of this class is Euler characteristic. –  Ryan Budney Aug 23 '11 at 0:57
A proof using Riemann--Roch and a Koszul-type resolution of the Diagonal can be found in the notes math.uni-bonn.de/people/hartmann/intersection_theory.pdf on my webpage. –  Heinrich Hartmann Aug 23 '11 at 10:57
@Heinrich: That's just perfect! I should come to you all the time with this kind of stuff. Seriously though, thanks a bunch. –  Jesko Hüttenhain Aug 25 '11 at 9:07

As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says $$\chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V)$$ where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes $${\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X)$$ (called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets $$\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX)$$ and by the Hodge decomposition theorem $$\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$$ where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.

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Do you use Poincare-Hopf in the proof of the general Gauss-Bonnet theorem? For example, there are common proofs of Gauss-Bonnet in the 2-dimensional (real) manifold case that use something that is very close to the Poincare-Hopf theorem -- the fact that the tangent vector to a closed curve in the plane has total winding number $\pm 1$. –  Ryan Budney Aug 24 '11 at 3:31
Another point of view is to use the (related) Lefschetz trace formula. The proof goes as follows: i) The Euler characterstic is the trace of the identity, hence is the intersection of the diagonal with itself in X x X. ii) The intersection of the diagonal with itself is computed by the excess intersection-formula, which gives the intersection in terms of the Chern class of the normal bundle of the diagonal embedding. iii) This is almost by definition the tangent or cotangent-bundle. –  D. Eriksson Aug 12 '14 at 11:12